THEORY OF STRUCTURES 2 CHAPTER 1 – APPROXIMATE ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES
Use of Approximate Methods When a model is used to represent any structure, the analysis of it must satisfy both the conditions of equilibrium and compatibility of displacement at the joints. As will be shown in later chapters of this text, the compatibility conditions for a statically statically indeterminate structure can be related to the loads provided we know the material’s modulus of elasticity and the size and shape of the members. For an initial design, however, we will not know a member’s size, and so a statically indeterminate analys is cannot be considered. For analysis a simpler model of the structure must be developed, one that is statically determinate. Once this model is specified, the analysis of it is called an approximate analysis. By performing an approximate analysis, a preliminary design of the member of structure can be made, and when this is complete, the more exact indeterminate analysis can then be performed and the design refined. An approximate analysis also provides insight as to a structure’s behavior under load and is i s beneficial when checking a more exact analysis or when time, money, or capability are not available for performing the more exact analysis. In a general sense, all methods of structural analysis are approximate, simply because the actual conditions of loading, geometry, material behavior, and joint resistance at the supports are never known in an exact sense. However, the statically indeterminate analysis of a structure will be referred to as an exact analysis, and the simpler statically determinate analysis will be referred to as the approximate analysis.
TRUSSES A common type of truss often used for lateral bracing of a building or for the top and bottom chords of a bridge is shown in Fig. 7-1a.
Fig. 7-1a
When used for this purpose, this truss is not considered a primary element for the support of the structure, and as a result is often analyzed by approximate methods. In the case shown, it will be noticed that if a diagonal is removed from each of the three panels, it will render the truss statically determinate. Hence, the truss is statically determinate to the third degree. Determinacy For any problem in truss analysis, it should be realized that the total number of unknowns includes the forces in “b” num ber of bars of the truss and the total number of external support reactions “r”. Since the truss members are all straight axial f orce members lying in the same plane, the force system acting at each joint is coplanar and concurrent. Consequently, rotational or moment equilibrium is automatically satisfied at the joint (or pin), and it is only necessary to satisfy ∑Fx = 0 and ∑Fy = 0 to ensure translational or force equilibrium. Therefore, only two equations of equilibrium can be written for each joint, and if there are “j” number of joints, the total number of equations available for solution is “2j”. By simply comparing the total number of unknowns “(b+r)” with the total number of available equilibrium equations, it is therefore possible to specify the determinacy for either a simple, compound, or complex truss. We have:
+ = 2 + > 2 In particular, the degree of indeterminacy is specified by the difference in the numbers:
+ − 2 Stability If:
+<2 Then the truss in unstable, that is, it will collapse, since there will be an insufficient number of bars or reactions to constrain all the joints. Also, a truss is unstable if it is statically determinate or statically indeterminate. In this case the stability will have to be determined either by inspection or by a force analysis.
From fig. 7-1a
=16,=3,=8 16+3>82 19>16 ℎ, 16+3 − 28 = 3 = And therefore we must take three assumptions regarding the bar forces in order to reduce the truss to one that is statically determinate. These assumptions can be made with regard to the cross-diagonals, realizing that when one diagonal in a panel is in tension the corresponding cross-diagonal will be in compression. This is evident from figure 7-1b
Fig. 7-1b
Where the “panel shear” V is carried by the vertical component of tensile forces in member “a” and the vertical component of compressive force in member “b”. Two methods of analysis are generally acceptable.
Method 1: If the diagonals are intentionally designed to be long and slender, it is reasonable to assume that they cannot support a compressive force; otherwise, they may easily buckle. Hence the panel shear is resisted entirely by the tension diagonal, whereas the compressive diagonal is assumed to be a zero-force member.
Method 2: If the diagonal members are intended to be constructed from large rolled sections such as angles or channels, they may be equally capable of supporting a tensile and compressive force. Here we will assume both tension and compression diagonals each carry half the panel shear.
Both of these methods of approximate analysis are illustrated numerically in the following examples.
EXAMPLES: 1. Determine (approximately) the forces in the members of the truss shown in fig. 7-2a. The diagonals are to be designed to support both tensile and compressive forces, and therefore each is assumed to carry half the panel shear.
SOLUTION:
=11,=3,=6 11+3>26 > ℎ, 11+3−26 = 2 =
∑ = 204 − 8 = 0 = + ∑= −−10−20+=0 −−10−20+10=0 = − (since negative, the force is upward)
−
=−= ℎ ℎ ,ℎ , = = ∑= 20−10−2(35) = 0 = . ℎ: = . = . ∑ = 8.33(45)3−3 = 0 = . ∑ = 8.33(45)3− 3 = 0 =.
∑ = −8.33(35)−10=0 = −
= = = ∑ = 10−2(35) = 0 = . = . = . ∑ = −8.33(45)3+ 3 = 0 =. ∑ = −8.33(45)3+ 3 = 0 =.
∑ = −.() =
= .
∑ = − + ().= = : Member FB AE FE AB AF DB EC ED BC DC EB
Force (kN) 8.33 8.33 6.67 6.67 15 8.33 8.33 6.67 6.67 5 10
Tension or Compression Tension Compression Compression Tension Tension Tension Compression Compression Tension Compression Tension
2. Determine (approximately) the forces in the members of the truss shown in figure. Assume the diagonals are slender and therefore will not support a compressive force.
=21,=3,=10 21+3>210
24>20 ℎ 21+3−210 = 4 =
∑ = −415 − 430 − 445 − 260 + 60 = 0 = ∑ = − 2 − 4 − 4 − 4 − 2 + = 0 − 2 − 4 − 4 − 4 − 2 + 8 = 0 =
−
= −= , = 0 ∑ = 0 8 − 2 − 45°=0 = . ∑ = 0 8.4945°15 − 15 = 0 = ∑ = − + 45°=0 −6+8.4945°=0 =
∑ = =
−
=8−2−4 =2 , = 0 ∑ = 8 − 2 − 4 − 45°=0 = . ∑ = 815 − 215 +2.8345°15− 15 = 0 = ∑ = 815 − 215 − 15 = 0 =
∑ = − +8.4945°=0 =
∑ = 22.8345° − = 0 = : Member AI, GE JB, DF JI, GF AB, DE JA, EF BH, DH IC, CG IH, HG BC, CD BI, DG CH
Force (k) 0 8.49 6 0 8 0 2.83 8 6 6 4
Tension or Compression Tension Compression Compression Tension Compression Tension Compression Compression
